Fractional View Analysis of Emden-Fowler Equations with the Help of Analytical Method

Abstract

This work aims at a new semi-analytical technique called the Adomian decomposition method for the analysis of time-fractional Emden–Fowler equations. The Laplace transformation and the iterative method are implemented to obtain the result of the given models. The suggested technique has the edge over other methods, as it does not need extra materials and calculations. The presented technique validity is demonstrated by examining four mathematical models. Due to the straightforward implementation, the proposed method can solve other non-linear fractional order problems.

1. Introduction

The study of fractional calculus (FC) and its several implementations in physics, mathematics, and applied sciences has received much attention in the twenty-first century. The dynamic device control theory, electrical networks, probability and statistics, optics, electrochemistry, signal processing and chemical mechanics are examples of FC applications. Fractional order partial differential equations are generalizations of classical partial differential equations. These have been of considerable interest in the recent literature [1,2,3]. These topics have received a great deal of attention, especially in electrochemical processes, viscoelasticity materials, colored noise, dielectric polarization, signal processing, anomalous diffusion, control theory and others. Increasingly, these models are used in fluid flow, finance and others. Most nonlinear fractional differential equations do not have analytic solutions, so approximation and numerical techniques must be used. The fractional movement Brownian, a Brownian motion generalization, is an important phenomenon in these calculation problems. Different definitions of fractional calculus can be found in a variety of articles and books [4,5,6,7,8,9]. Therefore, symmetry analysis is beautiful for studying partial differential equations, and especially when examining equations from the mathematical concepts of accounting. The key to nature is symmetry; however, the majority of natural observations lack symmetry. The phenomena of unexpected symmetry-breaking is an advanced mechanism for concealing symmetry. There are two types of symmetry: finite and infinitesimal. It is possible to have discrete or continuous finite symmetries. While space is a continuous change, parity and temporal inversion are discrete natural symmetries. Mathematicians have always been fascinated by patterns. The categorizing of spatial and planar patterns began in earnest during the eighteenth century. Unfortunately, solving fractional nonlinear differential equations precisely has proven to be quite difficult.

Numerous applied research fields, including mathematical biology, plasma physics, quantum mechanics, nonlinear optics, solid-state physics, chemical kinetics and fluid dynamics, etc., exhibit nonlinear effects. Various orders of nonlinear partial differential equations (PDEs) are used to simulate these processes. PDEs are frequently used to describe physical processes. The nonlinear nature of the majority of fundamental physical processes is concealed. The precise outcome of such nonlinear processes may be impossible to predict. This phenomenon can only be studied by employing tools to solve these nonlinear systems [10,11,12,13,14,15,16].

Fractional partial differential equations (FPDEs) are excellent for modeling various biological, economic, and dynamic phenomena. As a result, this topic has been the subject of much exciting and contemporary research during the past several decades. Additionally, FPDEs can be employed to simulate a vast array of phenomena, such as heat, sound, electromagnetic, electrodynamics, elastic, and hydrodynamics. See [17,18] for relevant research focusing on analytical and numerical solutions for FPDEs. The well-known FPDEs, such as the heat, wave, and Laplace equations, have been extensively studied, and there is a vast amount of literature on these subjects; see [19,20,21]. There, works [22,23,24,25,26,27,28] give reviews or/and developments of various numerical approaches to linear [22,23,24,25,26,27] and nonlinear [22,23,26] transport and diffusion problems, [29] introduces a homotopy perturbation method for nonlinear transport equations, ref. [30] proposes a perturbational approach to construct analytical approximations based on a double-parameter transformation perturbation expansion method, and finally the review paper [31] contains an exhaustive review of various modern fractional calculus applications.

The astrophysicists Jonathan Homer Lane and Robert Emden originally investigated the Lane–Emden equation when they looked at the thermal behavior of a spherical cloud of gas acting under the molecules’ attraction to one another and according to the classical principles of thermodynamics [32]. Since then, these singular equations defined by the Lane–Emden equations have been utilized in some applied science applications [33]. The Emden–Fowler model appears as a differential equation in astronomy and mathematical physics. The singularity behavior at the point $(\psi =0)$ makes it numerically challenging to solve the Emden–Fowler problem and other linear and nonlinear singular initial value problems in quantum mechanics and astronomy. This article uses the variational iteration transform approach to examine the approximation answer for the singular, linear and nonlinear fractional-order Emden–Fowler equation. The Emden–Fowler equation has been used to solve several numerical and analytical approaches, including the homotopy-perturbation technique [34], Haar wavelet collocation technique [35], residual power series technique [36], Sumudu transformation technique [37], and modified differential transform technique [38].

Adomain decomposition transform method is the mixture of two valuable methods, the Adomain decomposition technique and Laplace transformation introduced in [39,40]. Numerous physical phenomena which PDEs and PDEs represent are investigated by the Adomain decomposition transform method, such as the computational result of fractional Whitham–Broer–Kaup equations are discussed in [41], the result of linear and non linear fractional partial differential equations are effectively introduced in [42], the analytical result of the fractional nonlinear Volterra–Fredholm–Integro differential equations suggested in [43], and the schemes of delay differential equations are effectively discussed in [44]. The diffusion equation solution is given in [45] and the implementation of the proposed technique for other nonlinear PDEs can be mentioned in [46].

2. Definitions and Preliminaries Concepts

Definition 1.

The fractional integral Riemann–Liouville with order $\omega >0$ of function g is define as [47,48,49]

$$I_{\zeta }^{\omega }g\left(\zeta \right)=\left\{\begin{array}{cc}g\left(\zeta \right)\hfill & \mathrm{if}\omega =0\hfill \\ \frac{1}{\Gamma \left(\omega \right)}{\int }_0^{\zeta }{(\zeta -\upsilon )}^{\omega -1}g\left(\upsilon \right)d\upsilon \hfill & \mathrm{if}\omega >0,\hfill \end{array}\right.$$

Definition 2.

The fractional integral Caputo derivative with order $\omega \ge 0$ of function g is define as [47,48,49]

$$D^{\omega }g\left(\zeta \right)=\frac{{\partial }^{\omega }g\left(\zeta \right)}{\partial {\zeta }^{\omega }}=\left\{\begin{array}{cc}{I}^{n-\omega }\left[\frac{{\partial }^{\omega }g\left(\zeta \right)}{\partial {\zeta }^{\omega }}\right],\hfill & \mathrm{if}n-1<\omega \le n,n\in \mathbb{N}\hfill \\ \frac{{\partial }^{\omega }g\left(\zeta \right)}{\partial {\zeta }^{\omega }},\hfill \end{array}\right.$$

Lemma 1.

If $n-1<\omega \le n$ with $n\in \mathbb{N}$ and $g\in {\mathbb{C}}_{\Im }$ with $\zeta \ge -1,$ then [49,50]

$$\begin{array}{cc}& I^{\omega }{I}^{a}g\left(\zeta \right)=I^{\omega +a}g\left(\zeta \right),a,\omega \ge 0.\hfill \\ & I^{\omega }{\zeta }^{\lambda }=\frac{\Gamma (\lambda +1)}{\Gamma (\omega +\lambda +1)}{\zeta }^{\omega +\lambda },\omega >0,\lambda >-1,\zeta >0.\hfill \\ & I^{\omega }{D}^{\omega }g\left(\zeta \right)=g\left(\zeta \right)-\sum _{k=0}^{n-1}{g}^{\left(k\right)}\left(0^{+}\right)\frac{{\zeta }^k}{k!},for\zeta >0,n-1<\omega \le n.\hfill \end{array}$$

Definition 3.

The Laplace transformation of $g(\Im ),$$\Im >0$ is given as

$$G\left(s\right)=\mathbb{L}\left[g(\Im )\right]={\int }_0^{\infty }{e}^{-s\Im }g(\Im )d\Im .$$

Definition 4.

The convolution theorem for Laplace transform is given by

$$\mathbb{L}[g_1\ast g_2]=\mathbb{L}\left[g_1(\Im )\right]\ast \mathbb{L}\left[g_2(\Im )\right],$$

here, $g_1\ast g_2$ define the convolution between $g_1$ and $g_2$,

$$(g_1\ast g_2)(\Im )={\int }_0^{\Im }{g}_1\left(t\right)g_2(\Im -t)dt.$$

The fractional derivative of the Laplace transformation is

$$\mathbb{L}\left(D_{\Im }^{\omega }g(\Im )\right)=s^{\omega }\mathit{G}\left(\mathit{s}\right)-\sum _{k=0}^{n-1}{s}^{\omega -1-k}{g}^{\left(k\right)}\left(0\right),n-1<\omega <n,$$

where $G\left(s\right)$ is the Laplace transformation of $g(\Im )$.

Definition 5.

The Mittag–Leffler function, $E_{\omega }\left(p\right)$ for $\omega >0$ is defined as

$$E_{\omega }\left(p\right)=\sum _{n=0}^{\infty }\frac{p^n}{\Gamma \left(\omega n+1\right)}\omega >0,p\in \mathbb{C}.$$

3. Idea of the Laplace Adomian Decomposition Method (LADM)

In this section, the LADM is discussed for the solution of a fractional partial differential equation.

$$D^{\omega }u(\zeta ,\Im )+Lu(\zeta ,\Im )+Nu(\zeta ,\Im )=q(\zeta ,\Im ),\zeta \ge 0,\Im >0,0<\omega \le 1,$$

(1)

where $D^{\omega }=\frac{{\partial }^{\omega }}{\partial {\Im }^{\omega }}$ is the Caputo operator, and where L and N are linear and non-linear functions, q is the source function.

The initial condition is

$$u(\zeta ,0)=k\left(\zeta \right).$$

(2)

Using the Laplace transformation to Equation (1), we have

$$\mathbb{L}\left[D^{\omega }u(\zeta ,\Im )\right]+\mathbb{L}\left[Lu(\zeta ,\Im )+Nu(\zeta ,\Im )\right]=\mathbb{L}\left[q(\zeta ,\Im )\right],$$

(3)

and applying the Laplace transform differentiation property then $\omega \le 1$, we obtain

$$s^{\omega }\mathbb{L}\left[u(\zeta ,\Im )\right]-s^{\omega -1}u(\zeta ,0)=\mathbb{L}\left[q(\zeta ,\Im )\right]-\mathbb{L}\left[Lu(\zeta ,\Im )+Nu(\zeta ,\Im )\right],$$

$$\mathbb{L}\left[u(\zeta ,\Im )\right]=\frac{k\left(\zeta \right)}{s}+\frac{1}{s^{\omega }}\mathbb{L}\left[q(\zeta ,\Im )\right]-\frac{1}{s^{\omega }}\mathbb{L}\left[Lu(\zeta ,\Im )+Nu(\zeta ,\Im )\right].$$

(4)

The current method solution $u(\zeta ,\Im )$ is shown by the following infinite series

$$u(\zeta ,\Im )=\sum _{j=0}^{\infty }{u}_j(\zeta ,\Im ),$$

(5)

and the nonlinear terms (if any) in models are introduce by the infinite series of Adomian polynomials,

$$Nu(\zeta ,\Im )=\sum _{j=0}^{\infty }{A}_j,$$

(6)

$$A_j=\frac{1}{j!}{\left[\frac{d^j}{d{\lambda }^j}\left[N\sum _{j=0}^{\infty }\left({\lambda }^j{u}_j\right)\right]\right]}_{\lambda =0},j=0,1,2\cdots$$

(7)

substitution (5) and (7) in Equation (4), we obtain

$$\mathbb{L}\left[\sum _{j=0}^{\infty }u(\zeta ,\Im )\right]=\frac{k\left(\zeta \right)}{s}+\frac{1}{s^{\omega }}\mathbb{L}\left[q(\zeta ,\Im )\right]-\frac{1}{s^{\omega }}\mathbb{L}\left[L\sum _{j=0}^{\infty }{u}_j(\zeta ,\Im )+\sum _{j=0}^{\infty }{A}_j\right].$$

(8)

$$\begin{array}{c}\hfill \mathbb{L}\left[u_0(\zeta ,\Im )\right]=\frac{u(\zeta ,0)}{s}+\frac{1}{s^{\omega }}\mathbb{L}\left[q(\zeta ,\Im )\right],\\ \hfill \mathbb{L}\left[u_1(\zeta ,\Im )\right]=-\frac{1}{s^{\omega }}\mathbb{L}\left[L{u}_0(\zeta ,\Im )+A_0\right].\end{array}$$

(9)

Generally, we can write

$$\mathbb{L}\left[u_{j+1}(\zeta ,\Im )\right]=-\frac{1}{s^{\omega }}\mathbb{L}\left[L{u}_j(\zeta ,\Im )+A_j\right],j\ge 1.$$

(10)

Using the inverse Laplace transformation, in (10)

$$\begin{array}{cc}\hfill u_0(\zeta ,\Im )& ={\mathbb{L}}^{-1}\left\{\frac{u(\zeta ,0)}{s}+\frac{1}{s^{\omega }}\mathbb{L}\left[q(\zeta ,\Im )\right]\right\}\hfill \\ & =k\left(\zeta \right)+q(\zeta ,\Im )\frac{{\Im }^{\omega }}{\Gamma (\omega +1)}\hfill \end{array}$$

$$u_{j+1}(\zeta ,\Im )=-{\mathbb{L}}^{-1}\left[\frac{1}{s^{\omega }}\mathbb{L}\left[L{u}_j(\zeta ,\Im )+A_j\right]\right].$$

(11)

4. Results

Example 1. We consider the linear homogeneous time-fractional Emden–Fowler Heat equation [51]

$$\begin{array}{c}\hfill D_{\Im }^{\omega }u(\zeta ,\Im )=\frac{{\partial }^{2}u(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{5}{\zeta }\frac{\partial u(\zeta ,\Im )}{\partial \zeta }-(12{\Im }^2-2\Im {\zeta }^2+4{\Im }^4{\zeta }^2)u(\zeta ,\Im );0<\omega \le 1.\end{array}$$

(12)

with initial condition

$$u(\zeta ,0)=1.$$

(13)

subject to the Neumann and Dirichlet boundary conditions

$$\frac{\partial u(0,\Im )}{\partial \zeta }=0,u(1,\Im )=e^{{\Im }^2}.$$

The exact result at $\omega =1$ is given by

$$u(\zeta ,\Im )=e^{{\zeta }^2{\Im }^2}.$$

(14)

Solution Using LADM

The LADM begin by using Laplace transformation to (12) and applying the initial condition in (13), we obtain

$$\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[u(\zeta ,\Im )\right]=\sum _{k=0}^{\infty }{s}^{\omega -k-1}{u}^{\left(k\right)}(\zeta ,0)+{\mathbb{L}}_{\Im }\left[\frac{{\partial }^{2}u(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{5}{\zeta }\frac{\partial u(\zeta ,\Im )}{\partial \zeta }-(12{\Im }^2-2\Im {\zeta }^2+4{\Im }^4{\zeta }^2)u(\zeta ,\Im )\right].$$

(15)

Here, m = 1, so k = 0, then (15) becomes

$${\mathbb{L}}_{\Im }\left\{u(\zeta ,\Im )\right\}=\frac{1}{s}{u}^{\left(0\right)}(\zeta ,0)+\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left\{\frac{{\partial }^{2}u(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{5}{\zeta }\frac{\partial u(\zeta ,\Im )}{\partial \zeta }-(12{\Im }^2-2\Im {\zeta }^2+4{\Im }^4{\zeta }^2)u(\zeta ,\Im )\right\}.$$

(16)

In LADM the solution $u(\zeta ,\Im )$ is defined by infinite series as

$$u(\zeta ,\Im )=\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im ),$$

(17)

substituting (17) in (16), we get

$$\begin{array}{c}\hfill {\mathbb{L}}_{\Im }\left\{\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im )\right\}=\frac{1}{s}{u}^{\left(0\right)}(\zeta ,0)+\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }[\frac{{\partial }^2}{\partial {\zeta }^2}\left(\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im )\right)\\ \hfill +\frac{5}{\zeta }\frac{\partial }{\partial \zeta }\left(\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im )\right)-(12{\Im }^2-2\Im {\zeta }^2+4{\Im }^4{\zeta }^2)\left(\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im )\right)].\end{array}$$

(18)

By applying the linear property of the Laplace transformation, the following recursive relations are obtained:

$${\mathbb{L}}_{\Im }\left\{u_0(\zeta ,\Im )\right\}=\frac{1}{s}u(\zeta ,0),$$

(19)

$${\mathbb{L}}_{\Im }\left\{u_0(\zeta ,\Im )\right\}=\frac{1}{s}\left(1\right),$$

(20)

$${\mathbb{L}}_{\Im }\left\{u_1(\zeta ,\Im )\right\}=\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[\frac{{\partial }^2{u}_0(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{5}{\zeta }\frac{\partial u_0(\zeta ,\Im )}{\partial \zeta }-(12{\Im }^2-2\Im {\zeta }^2+4{\Im }^4{\zeta }^2)u_0(\zeta ,\Im )\right].$$

(21)

In general, for $n\ge 1$,

$${\mathbb{L}}_{\Im }\left\{u_{n+1}(\zeta ,\Im )\right\}=\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[\frac{{\partial }^2{u}_n(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{5}{\zeta }\frac{\partial u_n(\zeta ,\Im )}{\partial \zeta }-(12{\Im }^2-2\Im {\zeta }^2+4{\Im }^4{\zeta }^2)u_n(\zeta ,\Im )\right].$$

(22)

By applying an inverse Laplace transform, $u_n$$(n\ge 0)$ can be found as

$$\begin{array}{cc}\hfill u_0(\zeta ,\Im )& =1,\hfill \\ \hfill u_1(\zeta ,\Im )& =-\left[24\frac{{\Im }^{\omega +2}}{\Gamma (\omega +3)}-2{\zeta }^2\frac{{\Im }^{\omega +1}}{\Gamma (\omega +2)}+96{\zeta }^2\frac{{\Im }^{\omega +4}}{\Gamma (\omega +5)}\right],\hfill \\ \hfill u_2(\zeta ,\Im )& =-16\frac{{\Im }^{2\omega +1}}{\Gamma (2\omega +2)}+288\frac{\Gamma (\omega +5)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +4}}{\Gamma (2\omega +5)}-24\frac{\Gamma (\omega +4)}{\Gamma (\omega +2)}\frac{{\Im }^{2\omega +3}}{\Gamma (2\omega +4)}+1152{\zeta }^2\frac{\Gamma (\omega +7)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}\hfill \\ \hfill & -48{\zeta }^2\frac{\Gamma (\omega +4)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +3}}{\Gamma (2\omega +4)}+4{\zeta }^2\frac{\Gamma (\omega +3)}{\Gamma (\omega +2)}\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}-192{\zeta }^2\frac{\Gamma (\omega +6)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +}}{\Gamma (2\omega +6)}+96{\zeta }^2\frac{\Gamma (\omega +7)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}\hfill \\ \hfill & -8{\zeta }^4\frac{\Gamma (\omega +6)}{\Gamma (\omega +2)}\frac{{\Im }^{2\omega +5}}{\Gamma (2\omega +6)}+384{\zeta }^4\frac{\Gamma (\omega +9)}{\Gamma (\omega +4)}\frac{{\Im }^{2\omega +8}}{\Gamma (2\omega +9)},\hfill \end{array}$$

(23)

Hence, the solution in series form is

$$\begin{array}{c}\hfill u(\zeta ,\Im )=u_0+u_1+u_2+u_3+\cdots .\end{array}$$

(24)

$$\begin{array}{cc}\hfill & u(\zeta ,\Im )=1-\{24\frac{{\Im }^{\omega +2}}{\Gamma (\omega +3)}-2{\zeta }^2\frac{{\Im }^{\omega +1}}{\Gamma (\omega +2)}+96{\zeta }^2\frac{{\Im }^{\omega +4}}{\Gamma (\omega +5)}\}-16\frac{{\Im }^{2\omega +1}}{\Gamma (2\omega +2)}+288\frac{\Gamma (\omega +5)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +4}}{\Gamma (2\omega +5)}\hfill \\ \hfill & -24\frac{\Gamma (\omega +4)}{\Gamma (\omega +2)}\frac{{\Im }^{2\omega +3}}{\Gamma (2\omega +4)}+1152{\zeta }^2\frac{\Gamma (\omega +7)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}-48{\zeta }^2\frac{\Gamma (\omega +4)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +3}}{\Gamma (2\omega +4)}+4{\zeta }^2\frac{\Gamma (\omega +3)}{\Gamma (\omega +2)}\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}\hfill \\ \hfill & -192{\zeta }^2\frac{\Gamma (\omega +6)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +}}{\Gamma (2\omega +6)}+96{\zeta }^2\frac{\Gamma (\omega +7)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}-8{\zeta }^4\frac{\Gamma (\omega +6)}{\Gamma (\omega +2)}\frac{{\Im }^{2\omega +5}}{\Gamma (2\omega +6)}+384{\zeta }^4\frac{\Gamma (\omega +9)}{\Gamma (\omega +4)}\frac{{\Im }^{2\omega +8}}{\Gamma (2\omega +9)}+\cdots .\hfill \end{array}$$

(25)

Putting $\omega =1$ in (25), we obtain

$$u(\zeta ,\Im )=e^{{\zeta }^2{\Im }^2}.$$

(26)

The numerical simulation was carried out in order to determine whether the future algorithm would result in increased accuracy. Compared to the technique described, the future scheme provides amazing precision, as evidenced by the acquired findings. Figure 1 expressed the comparison of the analytical result with the exact result for Example 1 Figure 2, and shows the fractional order graphs of three and two dimensions.

Figure 1. The exact and approximate solutions figures of Example 1.

Figure 2. The different fractional-order graph at $\Im =0.5$ of Example 1.

Example 2. Consider the homogeneous time-fractional Emden–Fowler equation [51]

$$D_{\Im }^{\omega }u(\zeta ,\Im )=\frac{{\partial }^{2}u(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{2}{\zeta }\frac{\partial u(\zeta ,\Im )}{\partial \zeta }-(6+4{\zeta }^2-cos\Im )u(\zeta ,\Im ),0<\omega \le 1,$$

(27)

with initial condition

$$u(\zeta ,0)=e^{{\zeta }^2}.$$

(28)

The exact solution at $\omega =1$ is given by

$$u(\zeta ,\Im )=e^{{\zeta }^2+sin\Im },$$

(29)

Solution Using LADM

Applying inverse Laplace transform to (27) with initial condition (28), we obtain

$${\mathbb{L}}_{\Im }\left[u(\zeta ,\Im )\right]=\frac{1}{s}{e}^{{\zeta }^2}+\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }[\frac{{\partial }^{2}u(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{2}{\zeta }\frac{\partial u(\zeta ,\Im )}{\partial \zeta }-(6+4{\zeta }^2-cos\Im )u(\zeta ,\Im )].$$

(30)

In LADM, the solution is defined by infinite series as

$$u(\zeta ,\Im )=\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im ).$$

(31)

$$\begin{array}{c}\hfill {\mathbb{L}}_{\Im }\left[\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im )\right]=\frac{1}{s}{e}^{{\zeta }^2}+\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[\frac{{\partial }^2}{\partial {\zeta }^2}\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im )+\frac{2}{\zeta }\frac{\partial }{\partial \zeta }\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im )-(6+4{\zeta }^2-cos\Im )\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im )\right].\end{array}$$

(32)

By using the linearity property of Laplace transform the following recursive formula are obtained

$${\mathbb{L}}_{\Im }\left[u_0(\zeta ,\Im )\right]=\frac{1}{s}{e}^{{\zeta }^2}$$

(33)

$${\mathbb{L}}_{\Im }\left[u_1(\zeta ,\Im )\right]=\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[\frac{{\partial }^2{u}_0}{\partial {\zeta }^2}+\frac{2}{\zeta }\frac{\partial u_0}{\partial \zeta }-(6+4{\zeta }^2-cos\Im )u_0\right].$$

(34)

In general for $n\ge 1$, we have

$${\mathbb{L}}_{\Im }\left[u_{n+1}(\zeta ,\Im )\right]=\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[\frac{{\partial }^2{u}_n}{\partial {\zeta }^2}+\frac{2}{\zeta }\frac{\partial u_n}{\partial \zeta }-(6+4{\zeta }^2-cos\Im )u_n\right].$$

(35)

By applying an inverse Laplace transform, $u_n$$(n\ge 0)$, we obtain

$$\begin{array}{cc}\hfill & u_0(\zeta ,\Im )=e^{{\zeta }^2},\hfill \\ \hfill & u_1(\zeta ,\Im )=e^{{\zeta }^2}[\frac{{\Im }^{\omega }}{\Gamma (\omega +1)}-\frac{{\Im }^{\omega +2}}{\Gamma (\omega +3)}+\frac{{\Im }^{\omega +4}}{\Gamma (\omega +5)}-\frac{{\Im }^{\omega +6}}{\Gamma (\omega +7)}+\dots ],\hfill \\ \hfill & u_2(\zeta ,\Im )=e^{{\zeta }^2}\{[\frac{{\Im }^{2\omega }}{\Gamma (2\omega +1)}-\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}+\frac{{\Im }^{2\omega +4}}{\Gamma (2\omega +5)}-\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}+..]-\frac{1}{2!}[\frac{\Gamma (\omega +3)}{\Gamma (\omega +1)}\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}\hfill \\ \hfill & -\frac{\Gamma (\omega +5)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +4}}{\Gamma (2\omega +5)}+\frac{\Gamma (\omega +7)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}-\frac{\Gamma (\omega +9)}{\Gamma (\omega +7)}\frac{{\Im }^{2\omega +8}}{\Gamma (2\omega +9)}+\dots ]\hfill \\ \hfill & +\frac{1}{4!}\left[\frac{\Gamma (\omega +4)}{\Gamma (\omega +1)}\frac{{\Im }^{2\omega +4}}{\Gamma (2\omega +5)}-\frac{\Gamma (\omega +7)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}+\frac{\Gamma (\omega +9)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +8}}{\Gamma (2\omega +9)}-\frac{\Gamma (\omega +11)}{\Gamma (\omega +7)}\frac{{\Im }^{2\omega +10}}{\Gamma (2\omega +11)}+..\right]\hfill \\ \hfill & -\frac{1}{6!}\left[\frac{\Gamma (\omega +7)}{\Gamma (\omega +1)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}-\frac{\Gamma (\omega +9)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +8}}{\Gamma (2\omega +9)}+\frac{\Gamma (\omega +11)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +10}}{\Gamma (2\omega +11)}-\frac{\Gamma (\omega +13)}{\Gamma (\omega +7)}\frac{{\Im }^{2\omega +12}}{\Gamma (2\omega +13)}+\dots \right]+\dots \}\hfill \\ \hfill & .\hfill \\ \hfill & .\hfill \\ \hfill & .\hfill \end{array}$$

(36)

Hence, the series solution is given by $u(\zeta ,\Im )={\sum }_{n=0}^{\infty }$

$$u(\zeta ,\Im )=u_o(\zeta ,\Im )+u_1(\zeta ,\Im )+u_2(\zeta ,\Im )+\dots ,$$

(37)

$$\begin{array}{cc}\hfill & u(\zeta ,\Im )=e^{{\zeta }^2}\left[\frac{{\Im }^{\omega }}{\Gamma (\omega +1)}-\frac{{\Im }^{\omega +2}}{\Gamma (\omega +3)}+\frac{{\Im }^{\omega +4}}{\Gamma (\omega +5)}-\frac{{\Im }^{\omega +6}}{\Gamma (\omega +7)}+\cdots \right]\hfill \\ \hfill & +e^{{\zeta }^2}\{\left[\frac{{\Im }^{2\omega }}{\Gamma (2\omega +1)}-\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}+\frac{{\Im }^{2\omega +4}}{\Gamma (2\omega +5)}-\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}+\cdots \right]\hfill \\ \hfill & -\frac{1}{2!}\left[\frac{\Gamma (\omega +3)}{\Gamma (\omega +1)}\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}-\frac{\Gamma (\omega +5)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +4}}{\Gamma (2\omega +5)}+\frac{\Gamma (\omega +7)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}-\frac{\Gamma (\omega +9)}{\Gamma (\omega +7)}\frac{{\Im }^{2\omega +8}}{\Gamma (2\omega +9)}+\cdots \right]\hfill \\ \hfill & +\frac{1}{4!}\left[\frac{\Gamma (\omega +4)}{\Gamma (\omega +1)}\frac{{\Im }^{2\omega +4}}{\Gamma (2\omega +5)}-\frac{\Gamma (\omega +7)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}+\frac{\Gamma (\omega +9)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +8}}{\Gamma (2\omega +9)}-\frac{\Gamma (\omega +11)}{\Gamma (\omega +7)}\frac{{\Im }^{2\omega +10}}{\Gamma (2\omega +11)}+\cdots \right]\hfill \\ \hfill & -\frac{1}{6!}\left[\frac{\Gamma (\omega +7)}{\Gamma (\omega +1)}\frac{{\Im }^{2\omega +6}}{\Gamma (2\omega +7)}-\frac{\Gamma (\omega +9)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +8}}{\Gamma (2\omega +9)}+\frac{\Gamma (\omega +11)}{\Gamma (\omega +5)}\frac{{\Im }^{2\omega +10}}{\Gamma (2\omega +11)}-\frac{\Gamma (\omega +13)}{\Gamma (\omega +7)}\frac{{\Im }^{2\omega +12}}{\Gamma (2\omega +13)}+\cdots \right]+\cdots \}+\cdots ,\hfill \end{array}$$

(38)

Putting $\omega =1$ we get

$$u(\zeta ,\Im )=e^{{\zeta }^2+sin\Im }.$$

(39)

The numerical simulation was carried out in order to determine whether the future algorithm would result in increased accuracy. Compared to the technique described, the future scheme provides amazing precision, as evidenced by the acquired findings. Figure 3 expressed the comparison of the analytical result with the exact result for Example 2. Figure 4 shows the fractional order graphs of three and two dimensions.

Figure 3. The exact and approximate solution figures of Example 2.

Figure 4. The different fractional-order graph of $\omega$ at $\Im =0.5$ of Example 2.

Example 3. Consider the non-linear homogenous time-fractional Emden–Fowler equation [51]:

$$\begin{array}{c}\hfill D_{\Im \Im }^{\omega }u(\zeta ,\Im )=\frac{{\partial }^{2}u(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{6}{\zeta }\frac{\partial u(\zeta ,\Im )}{\partial \zeta }+(14\Im +{\zeta }^4)u+4\Im uln\left(u\right);1<\omega \le 2\end{array}$$

(40)

with initial condition

$$\begin{array}{c}\hfill u(\zeta ,0)=1,u_{\Im }(\zeta ,0)=-{\zeta }^2.\end{array}$$

(41)

The exact solution $\omega =1$ is given by

$$\begin{array}{c}\hfill u(\zeta ,\Im )=e^{-\Im {\zeta }^2}.\end{array}$$

(42)

Solution Using LADM

Applying Laplace transform to (40), we obtain

$$\begin{array}{cc}\hfill \frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[u(\zeta ,\Im )\right]& =\sum _{k=0}^{m-1}{s}^{\omega -k-1}{u}^k(\zeta ,0)+\sum _{k=0}^{m-1}{s}^{\omega -k-2}{u}_{\Im }^k(\zeta ,0)\hfill \\ \hfill & +{\mathbb{L}}_{\Im }\left[\frac{{\partial }^{2}u(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{6}{\zeta }\frac{\partial u(\zeta ,\Im )}{\partial \zeta }+(14\Im +{\zeta }^4)u+4\Im uln\left(u\right)\right].\hfill \end{array}$$

(43)

Here, $m=2$ so $k=1$, then

$$\begin{array}{c}\hfill {\mathbb{L}}_{\Im }\left[u(\zeta ,\Im )\right]=\frac{1}{s}u(\zeta ,0)+\frac{1}{s^2}{u}_{\Im }(\zeta ,0)+\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[\frac{{\partial }^{2}u(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{6}{\zeta }\frac{\partial u(\zeta ,\Im )}{\partial \zeta }+(14\Im +{\zeta }^4)u+4\Im uln\left(u\right)\right].\end{array}$$

(44)

In LADM, the solution $u(\zeta ,\Im )$ is defined by infinite series as

$$\begin{array}{c}\hfill u(\zeta ,\Im )=\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im ).\end{array}$$

(45)

The non-linear term

$$\begin{array}{c}\hfill N_1=uln\left(u\right),\end{array}$$

(46)

is generally expressed as an infinite series of the Adomian polynomial as

$$\begin{array}{c}\hfill N_1\left(u\right)=\sum _{n=0}^{\infty }{A}_n,\end{array}$$

(47)

where

$$\begin{array}{cc}\hfill A_n& =\frac{1}{n!}{\left[\frac{d^n}{d{\lambda }^n}\left[N_1\left(\sum _{k=0}^{\infty }{\lambda }^k{u}_k\right)\right]\right]}_{\lambda =0},\hfill \\ \hfill & =\frac{1}{n!}[\frac{d^n}{d{\lambda }^n}{\left[\left(\sum _{k=0}^{\infty }{\lambda }^k{u}_k\right)\left(ln\left(\sum _{k=0}^{\infty }{\lambda }^k{u}_k\right)\right)\right]}_{\lambda =0}.\hfill \end{array}$$

(48)

putting (45) and (47) in (44), we get

$$\begin{array}{cc}\hfill {\mathbb{L}}_{\Im }\left[\sum _{n=0}^{\infty }{u}_n(\zeta ,\Im )\right]& =\frac{1}{s}u(\zeta ,0)+\frac{1}{s^2}{u}_{\Im }(\zeta ,0)+\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }[\frac{{\partial }^2}{\partial {\zeta }^2}\left(\sum _{n=0}^{\infty }{u}_n\right)\hfill \\ \hfill & +\frac{6}{\zeta }\frac{\partial }{\partial \zeta }\left(\sum _{n=0}^{\infty }{u}_n\right)+(14\Im +{\zeta }^4)\left(\sum _{n=0}^{\infty }{u}_n\right)+4\Im \left(\sum _{n=0}^{\infty }{A}_n\right)].\hfill \end{array}$$

(49)

By using the linearity property of Laplace transform, the following recursive relation are obtained:

$$\begin{array}{cc}\hfill {\mathbb{L}}_{\Im }\left[u_0(\zeta ,\Im )\right]& =\frac{1}{s}u(\zeta ,0)+\frac{1}{s^2}{u}_{\Im }(\zeta ,0),\hfill \\ \hfill {\mathbb{L}}_{\Im }\left[u_1(\zeta ,\Im )\right]& =\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[\frac{{\partial }^2{u}_0}{\partial {\zeta }^2}+\frac{6}{\zeta }\frac{\partial u_0}{\partial \zeta }+(14\Im +{\zeta }^4)u_0+4\Im A_0\right].\hfill \end{array}$$

(50)

In general, for $n\ge 1$, we have

$$\begin{array}{cc}\hfill {\mathbb{L}}_{\Im }\left[u_{n+1}(\zeta ,\Im )\right]& =\frac{1}{s^{\omega }}{\mathbb{L}}_{\Im }\left[\frac{{\partial }^2{u}_{n+1}(\zeta ,\Im )}{\partial {\zeta }^2}+\frac{6}{\zeta }\frac{\partial u_n}{\partial \zeta }+(14\Im +{\zeta }^4)u_n+4\Im A_n\right].\hfill \end{array}$$

(51)

By applying an inverse Laplace transform, $u_n$$(n\ge 0)$ can be found as

$$\begin{array}{cc}\hfill & u_0(\zeta ,\Im )=1-\Im {\zeta }^2,\hfill \\ \hfill & u_1(\zeta ,\Im )=-28{\zeta }^2\frac{{\Im }^{\omega +2}}{\Gamma (\omega +3)}+{\zeta }^4\frac{{\Im }^{\omega }}{\Gamma (\omega +1)}-{\zeta }^6\frac{{\Im }^{\omega +1}}{\Gamma (\omega +2)},\hfill \\ \hfill & u_2(\zeta ,\Im )=-392\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}+36{\zeta }^2\frac{{\Im }^{2\omega }}{\Gamma (2\omega +1)}-66{\zeta }^4\frac{{\Im }^{2\omega +1}}{\Gamma (2\omega +2)}+392{\zeta }^2\frac{\Gamma (\omega +4)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +3}}{\Gamma (2\omega +4)}\hfill \\ \hfill & +14{\zeta }^4\frac{\Gamma (\omega +2)}{\Gamma (\omega +1)}\frac{{\Im }^{2\omega +1}}{\Gamma (2\omega +2)}-14{\zeta }^6\frac{\Gamma (\omega +3)}{\Gamma (\omega +2)}\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}-28{\zeta }^6\frac{2\omega +2}{\Gamma (2\omega +3)}+{\zeta }^8\frac{{\Im }^{2\omega }}{\gamma (2\omega +1)}+{\zeta }^{10}\frac{2\omega +1}{\Gamma (2\omega +2)},\hfill \\ \hfill & ⋮\hfill \end{array}$$

(52)

Hence, the series solution in series form is $u(\zeta ,\Im )={\sum }_{n=0}^{\infty }$

$$u(\zeta ,\Im )=u_0(\zeta ,\Im )+u_1(\zeta ,\Im )+u_2(\zeta ,\Im )+\cdots .$$

(53)

$$\begin{array}{cc}\hfill u(\zeta ,\Im )& =1-\Im {\zeta }^2-28{\zeta }^2\frac{{\Im }^{\omega +2}}{\Gamma (\omega +3)}+{\zeta }^4\frac{{\Im }^{\omega }}{\Gamma (\omega +1)}-{\zeta }^6\frac{{\Im }^{\omega +1}}{\Gamma (\omega +2)}\hfill \\ \hfill & -392\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}+36{\zeta }^2\frac{{\Im }^{2\omega }}{\Gamma (2\omega +1)}-66{\zeta }^4\frac{{\Im }^{2\omega +1}}{\Gamma (2\omega +2)}+392{\zeta }^2\frac{\Gamma (\omega +4)}{\Gamma (\omega +3)}\frac{{\Im }^{2\omega +3}}{\Gamma (2\omega +4)}+14{\zeta }^4\frac{\Gamma (\omega +2)}{\Gamma (\omega +1)}\frac{{\Im }^{2\omega +1}}{\Gamma (2\omega +2)}\hfill \\ \hfill & -14{\zeta }^6\frac{\Gamma (\omega +3)}{\Gamma (\omega +2)}\frac{{\Im }^{2\omega +2}}{\Gamma (2\omega +3)}-28{\zeta }^6\frac{2\omega +2}{\Gamma (2\omega +3)}+{\zeta }^8\frac{{\Im }^{2\omega }}{\gamma (2\omega +1)}+{\zeta }^{10}\frac{2\omega +1}{\Gamma (2\omega +2)}.\hfill \end{array}$$

(54)

The numerical simulation was carried out in order to determine whether the future algorithm would result in increased accuracy. In comparison to the technique described, the future scheme provides amazing precision, as evidenced by the acquired findings. Figure 5 expresses the comparison of the analytical result with the exact result for Example 3. Figure 6 shows the fractional order graphs of three and two dimensions. In Table 1, comparison of the exact and proposed method of different fractional order $\omega$.

Figure 5. The exact and approximate solutions figures of Example 3.

Figure 6. The different fractional-order graph of $\omega$ at $\Im =0.5$ for Example 3.

Table 1. Comparison of the exact and proposed technique solution and various fractional-orders $\omega$ for Example 3.

$\mathit{\zeta }=\mathit{\Im }$	LADM	LADM	LADM	LADM	LADM	LADM
	for $\mathit{\omega }=1.5$	for $\mathit{\omega }=1.6$	for $\mathit{\omega }=1.7$	for $\mathit{\omega }=1.8$	for $\mathit{\omega }=1.9$	for $\mathit{\omega }=2$
0	5.2714 $\times {10}^{-4}$	1.5764 $\times {10}^{-4}$	4.3971 $\times {10}^{-5}$	2.2429 $\times {10}^{-5}$	8.8187 $\times {10}^{-6}$	1.5425 $\times {10}^{-9}$
0.1	5.5938 $\times {10}^{-3}$	4.6680 $\times {10}^{-3}$	3.6036 $\times {10}^{-3}$	2.4014 $\times {10}^{-3}$	1.1397 $\times {10}^{-3}$	1.2773 $\times {10}^{-7}$
0.2	5.0322 $\times {10}^{-3}$	4.3098 $\times {10}^{-3}$	3.4931 $\times {10}^{-3}$	2.5330 $\times {10}^{-3}$	1.3759 $\times {10}^{-3}$	8.1484 $\times {10}^{-8}$
0.3	4.1904 $\times {10}^{-3}$	3.5627 $\times {10}^{-3}$	2.8775 $\times {10}^{-3}$	2.0985 $\times {10}^{-3}$	1.1662 $\times {10}^{-3}$	4.2659 $\times {10}^{-8}$
0.4	3.2784 $\times {10}^{-3}$	2.7336 $\times {10}^{-3}$	2.1622 $\times {10}^{-3}$	1.5449 $\times {10}^{-3}$	8.4562 $\times {10}^{-4}$	2.5738 $\times {10}^{-8}$
0.5	2.3462 $\times {10}^{-3}$	1.8956 $\times {10}^{-3}$	1.4464 $\times {10}^{-3}$	9.9290 $\times {10}^{-4}$	5.2127 $\times {10}^{-4}$	2.4648 $\times {10}^{-8}$
0.6	1.4372 $\times {10}^{-3}$	1.0967 $\times {10}^{-3}$	7.8131 $\times {10}^{-4}$	4.9376 $\times {10}^{-4}$	2.3525 $\times {10}^{-4}$	3.4371 $\times {10}^{-8}$
0.7	6.2014 $\times {10}^{-4}$	4.0001 $\times {10}^{-4}$	2.2073 $\times {10}^{-4}$	8.9498 $\times {10}^{-5}$	1.2319 $\times {10}^{-5}$	4.8842 $\times {10}^{-8}$
0.8	1.7881 $\times {10}^{-5}$	1.0509 $\times {10}^{-4}$	1.6401 $\times {10}^{-4}$	1.7263 $\times {10}^{-4}$	1.2137 $\times {10}^{-4}$	5.8531 $\times {10}^{-8}$
0.9	2.7151 $\times {10}^{-4}$	2.9046 $\times {10}^{-4}$	2.7740 $\times {10}^{-4}$	2.2816 $\times {10}^{-4}$	1.3775 $\times {10}^{-4}$	4.8794 $\times {10}^{-8}$
1	1.6263 $\times {10}^{-6}$	1.4268 $\times {10}^{-6}$	1.1378 $\times {10}^{-6}$	7.7845 $\times {10}^{-6}$	3.8059 $\times {10}^{-7}$	2.4304 $\times {10}^{-8}$

5. Conclusions

In this study, the Adomian decomposition transform technique is applied to fractional-order Emden–Fowler equations to yield analytical results. Using Adomian polynomials to solve fractional-order nonlinear partial differential equations makes the suggested method an effective and straightforward method for solving fractional-order partial differential equations. The present method can be used to various fractional-order partial differential equations, which commonly arise in applied mathematics, due to its modest amount of computations and clear and straightforward application. Moreover, the proposed technique provided the convergent series solutions with easily determinable components without using linearization, perturbation or limiting assumption. The analytical and graphical results achieved by the proposed method were computationally very attractive and more accurate to find the solutions of governing equation. In future work, we intend to extend the Laplace Adomain decomposition method to physical applications with higher dimensions and fuzzy differential equations.